geometrytrianglestrigonometry
An isosceles triangle $ABC$ is given $(AC=BC).$ The perimeter of $\triangle ABC$ is $2p$, and the base angle is $\alpha.$ Find the radius of the circumscribed circle $R$.
$$R=\frac{p}{2\sin\alpha(1+\cos\alpha)}$$
Let $CD=2R.$ The triangle $BCD$ is a right triangle and we have $\angle BAC=\angle ABC=\angle BDC=\alpha.$
I am not sure how to approach the problem. It's really hard for me to solve problems like this. Can you give me a hint and some thoughts on the problem?
See the figure below:
Let $I_1F = r_1$, $I_2G = r_2$, and $I_1E = r_3$.
Note that $GD =r_2$, $DF = r_1$, and $I_2E = r_3$. Note also that $\triangle I_1DI_2$ is a right triangle.
So using Pytagoras' Theorem in triangles $\triangle DGI_2$, $\triangle DFI_1$, and $\triangle I_1DI_2$ we get: $$DI_1= r_1 \sqrt{2},$$ $$DI_2= r_2 \sqrt{2},$$ $$I_1I_2= \sqrt{2(r_1^2+r_2^2)} \tag1$$
Let $m(\angle I_1AI_2)= \alpha$, as $AI_2$ is bisector of $\angle CAD$ and $AI_1$ is bisector of $\angle BAD$, we can conclude that $$\alpha = 45 ^{\circ}$$ But if $\alpha = 45 ^{\circ}$ then $\angle I_1EI_2$ is a right angle (central angle), and using Pytagoras' Theorem in $\triangle I_1EI_2$ and equation $(1)$ we get: $$r_3=\sqrt{r_1^2+r_2^2} \tag2$$
Now let's calculate $r$, the inscribed circle radius of $\triangle ABC$, and compare it with $r_3$.
See the picture below:
We know that $\triangle ADC$, $\triangle BDA$, and $\triangle BAC$ are similar, then $$\frac{r_1}{c}=\frac{r_2}{b}=\frac{r}{a}=k \tag3$$ So using relation $(3)$ and Pytagoras' Theorem again ($\triangle ABC$) we get: $$a^2=b^2+c^2 \Rightarrow \left(\frac{r}{k}\right)^2=\left(\frac{r_1}{k}\right)^2 + \left(\frac{r_2}{k}\right)^2 \Rightarrow $$ $$\Rightarrow r^2=r_1^2+r_2^2 \Rightarrow r=\sqrt{r_1^2+r_2^2} \tag4$$
Therefore comparing $(2)$ and $(4)$ we can conclude finally that $$r=r_3.$$
I saw the following solution may years ago:
On side $AD$ construct in exterior equilateral triangle $ADE$. Connect $BE$.
Then $AB=AC, AE=BC, \angle BAE=\angle ABC$ gives $\Delta BAE =\Delta ABC$ and hence $AB=BE$.
But then $$AB=BE, BD=BD, DA=DE \Rightarrow ADB =EDB$$
Hence $\angle ADB=\angle EDB$. Since the two angles add to $300^\circ$ they are each $150^\circ$. Then $\angle ABD + \angle ADB+ \angle BAD=180^\circ$ gives $ABD=10^\circ$.
Best Answer
Another simple approach. Let $x=AC=BC$. Then
$$2p=AC+BC+2AH\\=2x+2x\cos\alpha$$
and
$$R=\frac 12 CD=\frac 12 \frac{BC}{ \sin \alpha} = \frac{x}{2 \sin \alpha}$$
Now you can complete the solution by a simple substitution.